LESSON 5 POLYNOMIALS
Definition A polynomial p is a function of the form where
p( x )  a n x n  a n  1 x n  1  a n  2 x n  2       a 2 x 2  a 1 x  a 0 ,
where n is a positive integer. The a i ’s can be real or complex numbers for
i  0 , 1, 2 , 3 , . . . . , n and a n  0 .
NOTE: In this class, we will restrict our discussion to polynomials with real
(number) coefficients.
Notation: The coefficient a n is called the leading coefficient and n is called the
degree of the polynomial.
COMMENT: Any polynomial is a “continuous” function. The concept of
continuity will be defined in calculus. The continuity of a polynomial implies that
its graph can be drawn without lifting your pencil. Thus, there are no points
missing in the graph, and there are no jumps or breaks in the graph. In general, the
graph of any continuous function can be drawn without lifting your pencil.
COMMENT: The graph of any polynomial is “smooth.” The concept of smooth
will also be defined in calculus. The smoothness of the graph means that there are
no sharp turns in the graph.
Definition A real or complex number z is called a zero or a root of the polynomial
p if and only if p ( z )  0 .
Theorem If z  a  b i is a zero (root) of a polynomial with real coefficients,
then the conjugate z  a  b i is also a zero (root) of the polynomial.
Proof Recall the following properties for conjugates.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
z  w  z  w
1.
2.
By mathematical induction,
z1  z 2  z 3       z n = z1  z 2  z 3       z n
zw  z w
3.
4.
By mathematical induction, z 1 z 2 z 3     z n = z 1 z 2 z 3     z n
zn  z
5.
6.
n
If a is a real number, then a  a .
n
n 1
 a n  2 x n  2       a 2 x 2  a1 x  a 0 .
Let p( x )  a n x  a n  1 x
Since z is a zero (root) of the polynomial, then p ( z )  0 . We want to show that
p( z )  0 . p( z )  0 
a n z n  a n  1 z n  1  a n  2 z n  2       a 2 z 2  a1 z  a 0  0 
an z n  an  1 z n
1
 a n  2 z n  2       a 2 z 2  a1 z  a 0  0 
an z n  an  1 z n  1  an
 2
z n  2       a 2 z 2  a1 z  a 0  0 
a n z n  a n  1 z n  1  a n  2 z n  2       a 2 z 2  a1 z  a 0  0 
an z
n
 an
1
z
n 1
 an
 2
z
n  2
      a2 z
2
 a1 z  a 0  0 
p( z )  0 . Thus, z is a zero (root) of the polynomial.
Note, since the coefficients of the polynomial are real numbers, then a i  a i for
i  0 , 1, 2 , 3 , . . . . , n .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
NOTE: This proof is easier to read (and type) if we make use of summation
notation
p( x )  a n x n  a n  1 x n  1  a n  2 x n  2       a 2 x 2  a 1 x  a 0 =
n

i  0
n

i  0
n
a i x i . Then p( z )  0 
ai z i  0 
n

i  0

i  0
n
a
ai z i  0 
ai z i  0 
i  0
n

i  0
ai z
i
i
zi  0 
 0  p( z )  0 .
Theorem Let a be a real number. Let p be any polynomial. Then the following
statements are equivalent.
1.
x  a is a zero (root) of the polynomial p
2.
p( a )  0
3.
x  a is a factor of p ( x )
4.
( a , 0 ) is an x-intercept of the graph of y  p( x )
Definition Let p be any polynomial. If ( x  z ) m , where m is positive integer,
is a factor of p ( x ) , then x  z is called a zero (root) of multiplicity m.
Theorem Let p be any polynomial. Let a be a real number. If ( x  a ) m is a
factor of p ( x ) , then
1.
if m is odd, then the graph of y  p( x ) crosses the x-axis at the x-intercept
(a, 0) ,
2.
if m is even, then the graph of y  p( x ) touches the x-axis at the
x-intercept ( a , 0 ) but does not cross the x-axis.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
Theorem (The Fundamental Theorem of Algebra) An nth degree polynomial has
at most n zeros (roots). If you count the multiplicity of the zero (root), then an
nth degree polynomial has exactly n zeros (roots).
Theorem An nth degree polynomial has at most n  1 local extremum points
(turning points).
Information about local extremum points can be obtained using calculus.
Theorem If a and b are real zeros (roots) of a polynomial and a  b , then the
polynomial has at least one local extremum point whose x-coordinate is between a
and b.
Now, we will look at the behavior of a polynomial for numerically large numbers.
Recall that the mathematical symbol  means implies. For example,
x 2  3  x   3 . That is, x 2  3 implies x is positive square root of 3
or x is negative square root of 3.
The mathematical symbol  means approaches. For example, x  5 means
that x approaches 5.
Real Number Line:


Then x   means x approaches positive infinity. That is, x grows positively
without bound. Also, x    means x approaches negative infinity. That is,
x grows negatively without bound.
Now, consider
p( x )  a n x n  a n  1 x n  1  a n  2 x n  2       a 2 x 2  a 1 x  a 0
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
for numerically large values of x positive or negative. In order to do this, we will
need to rewrite p ( x ) . Factoring out x n , we obtain that
 an x n
an  1 x n  1
an  2 x n  2
a2 x 2
a1 x
a0 

p( x )  x 










n
n
n
n
n  =
 xn
x
x
x
x
x


n
an  1
an  2
a2
a1
a0 


x n  a n 









2
n  2
n  1
n 
x
x
x
x
x


As x   or x    , we have that
a2
xn
 2
 0,
a1
xn
 1
 0 , and
a0
xn
an
 1
x
 0,
an
x
 2
2
 0, . . . . ,
 0.
n
Thus, we have that p( x )  a n x for numerically large values of x positive or
negative.
  , a n  0
p
(
x
)


Thus, as x   ,
   , a n  0



p
(
x
)


and as x    ,



  , n is odd and a n  0
  , n is even and a n  0
 , n is odd and a n  0
 , n is even and a n  0
Theorem (Leading Coefficient) If the leading coefficient of a polynomial p is
positive, then p( x )   as x   . If the leading coefficient of a polynomial
p is negative, then p( x )    as x   . That is, the sign of the leading
coefficient tells you the sign of the infinity that p ( x ) approaches as x   .
Copyrighted by James D. Anderson, The University of Toledo
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Theorem Let p be a polynomial whose degree is even.
1.
If p( x )   as x   , then p( x )   as x    .
2.
If p( x )    as x   , then p( x )    as x    .
That is, for an even degree polynomial p, p ( x ) approaches the same signed
infinity as x   and x    .
Theorem Let p be a polynomial whose degree is odd.
1.
If p( x )   as x   , then p( x )    as x    .
2.
If p( x )    as x   , then p( x )   as x    .
That is, for an odd degree polynomial p, p ( x ) approaches different signed
infinities as x   and x    .
Definition A function f is said to be even if and only if f (  x )  f ( x ) for all
x in the domain of f. A function f is said to be odd if and only if
f (  x )   f ( x ) for all x in the domain of f.
COMMENT: The graph of an even function is symmetry about the y-axis, and the
graph of an odd function is symmetry through the origin.
Examples Find the zeros (roots) and their multiplicities. Discuss the implication
of the multiplicity on the graph of the polynomial. Determine the sign of the
infinity that the polynomial values approaches as x or t approaches positive
infinity and negative infinity. Use this information to determine the number of
local extremum points (turning points) that the graph of the polynomial has.
Determine whether the polynomial is even, odd, or neither in order to make use of
symmetry if possible.
1.
f ( x )  2 x 3  72 x
Zeros (Roots) of f : f ( x )  0  2 x 3  72 x  0 
2 x ( x 2  36 )  0  2 x ( x  6 ) ( x  6 )  0  x  0 , x   6 ,
x  6
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
Since the factor x produces the zero (root) of 0, its multiplicity is one.
Since the factor x  6 produces the zero (root) of  6 , its multiplicity is
one. Since the factor x  6 produces the zero (root) of 6, its multiplicity is
one.
Zero (Root)
Multiplicity
Implication for the Graph
of the Polynomial
6
1
Crosses the x-axis at (  6 , 0 )
0
1
Crosses the x-axis at ( 0 , 0 )
6
1
Crosses the x-axis at ( 6 , 0 )
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
f ( x )  2 x 3  72 x is 2 > 0, then f ( x )   as x   .
x   :
The degree of f ( x )  2 x 3  72 x is 3, which is odd. Since f ( x )  
as x   , then f ( x )    as x    .
The polynomial is odd: f ( x )  2 x 3  72 x 
f (  x )  2 (  x ) 3  72 (  x )   2 x 3  72 x   ( 2 x 3  72 x )   f ( x )
Thus, the graph of f is symmetry through the origin.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
y
y
x
x
The Drawing of this Sketch
Since the degree of the polynomial is 3, then the graph of the polynomial can
have at most 2 local extremum points (turning points).
The graph of the continuous polynomial must cross the x-axis at the point
(  6 , 0 ) . Then the graph must cross the x-axis at the origin. In order for this
to happen, there must be a local extremum point (turning point) whose xcoordinate is between  6 and 0. Then the graph must cross the x-axis at
the point ( 6 , 0 ) . In order for this to happen, there must be a local extremum
point (turning point) whose x-coordinate is between 0 and 6. We would
need calculus in order to obtain information about the x-coordinate of these
two local extremum points.
Thus, the graph of the polynomial has two local extremum points (turning
points).
2.
g( x )  6 x 3  7 x 2
Zeros (Roots) of g : g ( x )  0  6 x 3  7 x 2  0 
x 2 (6 x  7)  0  x  0 , x 
7
6
Copyrighted by James D. Anderson, The University of Toledo
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Since the factor x 2 produces the zero (root) of 0, its multiplicity is two.
7
Since the factor 6 x  7 produces the zero (root) of
, its multiplicity is
6
one.
Multiplicity
Implication for the Graph
of the Polynomial
0
2
Touches the x-axis at ( 0 , 0 )
7
6
1
Crosses the x-axis at
Zero (Root)
7

 ,0
6

Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
g ( x )  6 x 3  7 x 2 is 6 > 0, then g ( x )   as x   .
x   :
The degree of g ( x )  6 x 3  7 x 2 is 3, which is odd. Since g ( x )  
as x   , then g ( x )    as x    .
The polynomial is neither even nor odd: g ( x )  6 x 3  7 x 2 
g(  x )  6 (  x ) 3  7 (  x ) 2   6 x 3  7 x 2
Thus, g (  x )  g ( x ) and g (  x )   g ( x ) .
Thus, the graph of the polynomial is not symmetry with respect to the y-axis
nor the origin.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
y
y
7
6
7
6
x
x
The Drawing of this Sketch
Since the degree of the polynomial is 3, then the graph of the polynomial can
have at most 2 local extremum points (turning points).
Since the graph touches at the origin, then there is a local extremum point
(turning point) at the origin. Since the graph of the continuous polynomial
must cross the x-axis at the point  7 , 0  , then there is another local
6

extremum point (turning point) whose x-coordinate is between 0 and
7
6
. We
would need calculus in order to obtain information about this x-coordinate.
Thus, the graph of the polynomial has two local extremum points (turning
points).
3.
h( x )  48  32 x  3 x 2  2 x 3
To solve for the zeros (roots) of this polynomial, you we need to recall a
technique of factoring called “factor by grouping.”
Zeros (Roots) of h : h( x )  0  48  32 x  3 x 2  2 x 3  0 
16 ( 3  2 x )  x 2 ( 3  2 x )  0  ( 3  2 x ) ( 16  x 2 )  0 
Copyrighted by James D. Anderson, The University of Toledo
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(3  2 x )(4  x )(4  x )  0  x  
3
, x   4, x  4
2
3
Since the factor 3  2 x produces the zero (root) of  , its multiplicity is
2
one. Since the factor 4  x produces the zero (root) of  4 , its
multiplicity is one. Since the factor 4  x produces the zero (root) of 4,
its multiplicity is one.
Zero (Root)
Multiplicity
Implication for the Graph
of the Polynomial
4
1
Crosses the x-axis at (  4 , 0 )
3
2
1
Crosses the x-axis at
1
Crosses the x-axis at ( 4 , 0 )

4
 3 
 ,0
 2 
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
h( x )  48  32 x  3 x 2  2 x 3 is  2  0 , then h( x )    as
x  .
x   :
The degree of h( x )  48  32 x  3 x 2  2 x 3 is 3, which is odd. Since
h( x )    as x   , then h( x )   as x    .
The polynomial is neither even nor odd:
h( x )  48  32 x  3 x 2  2 x 3 
h(  x )  48  32 (  x )  3 (  x ) 2  2 (  x ) 3 =
Copyrighted by James D. Anderson, The University of Toledo
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48  32 x  3 x 2  2 x 3
Thus, h(  x )  h( x ) and h(  x )   h( x ) .
Thus, the graph of the polynomial is not symmetry with respect to the y-axis
nor the origin.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
y

3
2
y
x

3
2
x
The Drawing of this Sketch
Since the degree of the polynomial is 3, then the graph of the polynomial can
have at most 2 local extremum points (turning points).
The graph of the continuous polynomial must cross the x-axis at the point
(  4 , 0 ) . Then the graph must cross the x-axis at the point   3 , 0  . In

2

order for this to happen, there must be a local extremum point (turning point)
whose x-coordinate is between  4 and  3 . Then the graph must cross the
2
x-axis at the point ( 4 , 0 ) . In order for this to happen, there must be a local
extremum point (turning point) whose x-coordinate is between  3 and 4.
2
Copyrighted by James D. Anderson, The University of Toledo
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We would need calculus in order to obtain information about the x-coordinate
of these two local extremum points.
Thus, the graph of the polynomial has two local extremum points (turning
points).
4.
p( x )  5 x 3  20 x
Zeros (Roots) of p : p( x )  0  5 x 3  20 x  0 
5x( x 2  4)  0 
x  0 , x   2i
NOTE: x 2  4  0  x 2   4  x  
 4   2i
Since the factor x produces the zero (root) of 0, its multiplicity is one.
Since the factor x 2  4 produces the zeros (roots) of  2 i , the multiplicity
of each zero (root) is one.
Multiplicity
Implication for the Graph
of the Polynomial
0
1
Crosses the x-axis at ( 0 , 0 )
 2i
1
No implication
2i
1
No implication
Zero (Root)
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
p( x )  5 x 3  20 x is 5 > 0, then p( x )   as x   .
x   :
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
The degree of p( x )  5 x 3  20 x is 3, which is odd. Since p( x )  
as x   , then p( x )    as x    .
The polynomial is odd: p( x )  5 x 3  20 x 
p(  x )  5 (  x ) 3  20 (  x )   5x 3  20 x   ( 5x 3  20 x )   p( x )
Thus, the graph of p is symmetry through the origin.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
y
y
x
x
The Drawing of this Sketch
Since the degree of the polynomial is 3, then the graph of the polynomial can
have at most 2 local extremum points (turning points).
The graph of the continuous polynomial must cross the x-axis at the origin. If
the graph had a local extremum point (turning point) for some x  0 , then it
would have to have a second one in order to pass through the origin. Because
the graph is symmetry through the origin, the graph would have to have two
local extremum points (turning points) for x  0 . Thus, the polynomial
would have four local extremum points (turning points).
Copyrighted by James D. Anderson, The University of Toledo
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Thus, the graph of the polynomial has no local extremum points (turning
points).
5.
q( x )  54  45 x  12 x 2  x 3
This polynomial can not be factored using the Factor by Grouping technique.
In Lesson 7, we will learn a technique for factoring this polynomial. At that
point we will discover that
54  45 x  12 x 2  x 3  ( 6  x ) ( x  3 ) 2
Zeros (Roots) of q : q( x )  0  54  45 x  12 x 2  x 3  0 
( 6  x ) ( x  3) 2  0  x  6 , x  3
Since the factor 6  x produces the zero (root) of 6, its multiplicity is one.
Since the factor ( x  3 ) 2 produces the zero (root) of 3, its multiplicity is
two.
Multiplicity
Implication for the Graph
of the Polynomial
3
2
Touches the x-axis at ( 3 , 0 )
6
1
Crosses the x-axis at ( 6 , 0 )
Zero (Root)
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
q( x )  54  45 x  12 x 2  x 3 is  1  0 , then q( x )    as
x  .
x   :
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
The degree of q( x )  54  45 x  12 x 2  x 3 is 3, which is odd. Since
q( x )    as x   , then q( x )   as x    .
The polynomial is neither even nor odd. Thus, the graph of the polynomial is
not symmetry with respect to the y-axis nor the origin.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
y
y
x
x
The Drawing of this Sketch
Since the degree of the polynomial is 3, then the graph of the polynomial can
have at most 2 local extremum points (turning points).
Since the graph touches at the point ( 3 , 0 ) , then there is a local extremum
point (turning point) at this point. Since the graph of the continuous
polynomial must cross the x-axis at the point ( 6 , 0 ) , then there is another
local extremum point (turning point) whose x-coordinate is between 3 and 6.
We would need calculus in order to obtain information about this xcoordinate.
Thus, the graph of the polynomial has two local extremum points (turning
points).
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
6.
q( t )  t 3  16 t 2  64 t
For this problem, it will be helpful to recall the following special factoring
formula:
a 2  2ab  b 2  (a  b) 2
Of course, we also have that a 2  2 a b  b 2  ( a  b ) 2
Zeros (Roots) of q : q( t )  0  t 3  16 t 2  64 t  0 
t ( t 2  16 t  64 )  0  t ( t  8 ) 2  0  t  0 , t   8
Since the factor t produces the zero (root) of 0, its multiplicity is one. Since
the factor ( t  8 ) 2 produces the zero (root) of  8 , its multiplicity is two.
Multiplicity
Implication for the Graph
of the Polynomial
0
1
Crosses the t-axis at ( 0 , 0 )
8
2
Touches the t-axis at (  8 , 0 )
Zero (Root)
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the t-axis, and if the multiplicity of the zero (root) is even,
the graph of the polynomial touches the t-axis.
t  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
q( t )  t 3  16 t 2  64 t is 1 > 0, then q( t )   as t   .
t   :
The degree of q( t )  t 3  16 t 2  64 t is 3, which is odd.
q( t )   as t   , then q( t )    as t    .
Since
The polynomial is neither even nor odd. Thus, the graph of the polynomial is
not symmetry with respect to the y-axis nor the origin.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as t   and as t    .
y
y
t
t
The Drawing of this Sketch
Since the degree of the polynomial is 3, then the graph of the polynomial can
have at most 2 local extremum points (turning points).
Since the graph touches at the point (  8 , 0 ) , then there is a local extremum
point (turning point) at this point. Since the graph of the continuous
polynomial must cross the t-axis at the origin, then there is another local
extremum point (turning point) whose t-coordinate is between  8 and 0.
We would need calculus in order to obtain information about this tcoordinate.
Thus, the graph of the polynomial has two local extremum points (turning
points).
7.
f ( x )  x 3  2 x 2  9 x  18
NOTE: The expression
grouping.
x 3  2 x 2  9 x  18
can be factored by
Zeros (Roots) of f : f ( x )  0  x 3  2 x 2  9 x  18  0 
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
x 2 ( x  2)  9( x  2)  0  ( x  2)( x 2  9)  0 
x  2 , x   3i
Since the factor x  2 produces the zero (root) of 2, its multiplicity is one.
Since the factor x 2  9 produces the zeros (roots) of  3 i , the multiplicity
of each zero (root) is one.
Multiplicity
Implication for the Graph
of the Polynomial
2
1
Crosses the x-axis at ( 2 , 0 )
 3i
1
No implication
3i
1
No implication
Zero (Root)
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
f ( x )  x 3  2 x 2  9 x  18 is 1 > 0, then f ( x )   as x   .
x   :
The degree of f ( x )  x 3  2 x 2  9 x  18 is 3, which is odd. Since
f ( x )   as x   , then h( x )    as x    .
The polynomial is neither even nor odd. Thus, the graph of the polynomial is
not symmetry with respect to the y-axis nor the origin.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
y
x
Since the degree of the polynomial is 3, then the graph of the polynomial can
have at most 2 local extremum points (turning points).
The graph of the continuous polynomial could have no local extremum points
(turning points) or it could have two local extremum points (turning points).
There is not enough information from the zeros (roots) and the graph is not
symmetry to the y-axis nor the origin. We need calculus in order to get
information about the local extremum point(s).
8.
g ( x )  3 x 4  5 x 3  12 x 2
Zeros (Roots) of g : g ( x )  0  3 x 4  5 x 3  12 x 2  0 
x 2 ( 3 x 2  5 x  12 )  0  x 2 ( x  3 ) ( 3 x  4 )  0 
x  0, x   3 x 
4
3
Since the factor x 2 produces the zero (root) of 0, its multiplicity is two.
Since the factor x  3 produces the zero (root) of  3 , its multiplicity is
one. Since the factor 3 x  4 produces the zero (root) of 4 , its multiplicity
3
is one.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
Zero (Root)
Multiplicity
Implication for the Graph
of the Polynomial
3
1
Crosses the x-axis at (  3 , 0 )
0
2
Touches the x-axis at ( 0 , 0 )
4
3
1
Crosses the x-axis at
4

 ,0
3


Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
g ( x )  3 x 4  5 x 3  12 x 2 is 3 > 0, then g ( x )   as x   .
x   :
The degree of g ( x )  3 x 4  5 x 3  12 x 2 is 4, which is even. Since
g ( x )   as x   , then g ( x )   as x    .
The polynomial is neither even nor odd. Thus, the graph of the polynomial is
not symmetry with respect to the y-axis nor the origin.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
y
y
4
3
4
3
x
x
The Drawing of this Sketch
Since the degree of the polynomial is 4, then the graph of the polynomial can
have at most 3 local extremum points (turning points).
The graph of the continuous polynomial must cross the x-axis at the point
(  3 , 0 ) . Then the graph must touch the x-axis at the origin. In order for
this to happen, there must be a local extremum point (turning point) whose xcoordinate is between  3 and 0. Since the graph touches at the origin,
then there is a local extremum point (turning point) at the origin. Then the
graph of the polynomial must cross the x-axis at the point  4 , 0  . In order
 3

for this to happen, there must be a local extremum point (turning point)
whose x-coordinate is between 0 and 4 . We would need calculus in order
3
to obtain information about the x-coordinate of two of these local extremum
points (turning points).
Thus, the graph of the polynomial has three local extremum points (turning
points).
9.
h( x )  x 4  29 x 2  100
NOTE: The expression x 4  29 x 2  100 is quadratic in x 2 . Thus, it
factors like a 2  29 a  100 , where a  x 2 . Since a 2  29 a  100 =
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
( a  4 ) ( a  25 ) , then x 4  29 x 2  100  ( x 2  4 ) ( x 2  25 ) .
Zeros (Roots) of h : h( x )  0  x 4  29 x 2  100  0 
( x 2  4 ) ( x 2  25 )  0  x   2 , x   5
Since the factor x 2  4 produces the zeros (roots) of  2 and 2, the
multiplicity of each zero (root) is one. Since the factor x 2  25 produces
the zeros (roots) of  5 and 5, the multiplicity of each zero (root) is one.
Zero (Root)
Multiplicity
Implication for the Graph
of the Polynomial
5
1
Crosses the x-axis at (  5 , 0 )
2
1
Crosses the x-axis at (  2 , 0 )
2
1
Crosses the x-axis at ( 2 , 0 )
5
1
Crosses the x-axis at ( 5 , 0 )
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
h( x )  x 4  29 x 2  100 is 1 > 0, then h( x )   as x   .
x   :
The degree of h( x )  x 4  29 x 2  100 is 4, which is even. Since
h( x )   as x   , then h( x )   as x    .
The polynomial is even. Thus, the graph of h is symmetry about the y-axis.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
y
y
x
x
The Drawing of this Sketch
Since the degree of the polynomial is 4, then the graph of the polynomial can
have at most 3 local extremum points (turning points).
The graph of the continuous polynomial must cross the x-axis at the point
(  5 , 0 ) . Then the graph must cross the x-axis at the point (  2 , 0 ) . In
order for this to happen, there must be a local extremum point (turning point)
whose x-coordinate is between  5 and  2 . Then the graph must cross
the x-axis at the point ( 2 , 0 ) . In order for this to happen, there must be a
local extremum point (turning point) whose x-coordinate is between  2
and 2. Then the graph must cross the x-axis at the point ( 5 , 0 ) . In order for
this to happen, there must be a local extremum point (turning point) whose xcoordinate is between 2 and 5. We would need calculus in order to obtain
information about the x-coordinate of these three local extremum points.
Thus, the graph of the polynomial has three local extremum points (turning
points).
10.
f ( x )   16 x 4  56 x 2  49
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
The expression  16 x 4  56 x 2  49 is quadratic in x 2 . Thus, it factors
like  16 u 2  56 u  49 , where u  x 2 . Since  16 u 2  56 u  49 =
 ( 16 u 2  56 u  49 ) =  ( 4 u  7 ) 2 , then  16 x 4  56 x 2  49 =
 (4 x 2  7) 2
NOTE: We used the special factoring formula a 2  2 a b  b 2  ( a  b ) 2
to factor 16 u 2  56 u  49 since 16 u 2  56 u  49 =
( 4 u ) 2  2 ( 28 u )  7 2 .
Zeros (Roots) of f : f ( x )  0   16 x 4  56 x 2  49  0 
 (4 x 2  7) 2  0  4 x 2  7  0  x  
7
2
Since the factor ( 4 x 2  7 ) 2 produces the zeros (roots) of

7
2
and
7
2
,
the multiplicity of each zero (root) is two.
Zero (Root)

7
2
7
2
Multiplicity
Implication for the Graph
of the Polynomial
2
Touches the x-axis at


7

,0


2


2
Touches the x-axis at
 7


,0
 2



Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
f ( x )   16 x 4  56 x 2  49 is  16  0 , then f ( x )    as x   .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
x   :
The degree of f ( x )   16 x 4  56 x 2  49 is 4, which is even. Since
f ( x )    as x   , then f ( x )    as x    .
The polynomial is even. Thus, the graph of f is symmetry about the y-axis.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
y

7
2
y
7
2
x
7
2

7
2
x
The Drawing of this Sketch
Since the degree of the polynomial is 4, then the graph of the polynomial can
have at most 3 local extremum points (turning points).
Since the graph of the continuous polynomial touches the x-axis at the point


7

, 0 ,


2


then there is a local extremum point (turning point) at this point.
Then the graph must touch the x-axis at the point
 7


, 0 .
 2



In order for this
to happen, there must be a local extremum point (turning point) whose xcoordinate is between

7
2
and
7
2
. Since the graph touches at the point
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
 7


, 0 ,
 2



then there is a local extremum point (turning point) at this point.
We would need calculus in order to obtain information about the x-coordinate
of the local extremum points (turning points) whose x-coordinate is between

7
2
and
7
2
.
Thus, the graph of the polynomial has three local extremum points (turning
points).
11.
p( x )  15  2 x 2  x 4
NOTE: The expression 15  2 x 2  x 4 is quadratic in x 2 . Thus, it
factors like 15  2 a  a 2 , where a  x 2 . Since 15  2 a  a 2 =
( 5  a ) ( 3  a ) , then 15  2 x 2  x 4  ( 5  x 2 ) ( 3  x 2 ) .
Zeros (Roots) of h : p( x )  0  15  2 x 2  x 4  0 
(5  x 2 )(3  x 2 )  0  x   i
5, x  
3
Since the factor 5  x 2 produces the zeros (roots) of  i
5 and i 5 ,
Since the factor 3  x 2
the multiplicity of each zero (root) is one.
3 , the multiplicity of each zero
produces the zeros (roots) of  3 and
(root) is one.
Zero (Root)

3
3
i
i
5
5
Multiplicity
Implication for the Graph
of the Polynomial
1
Crosses the x-axis at ( 
1
Crosses the x-axis at ( 3 , 0 )
1
No implication
1
No implication
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
3 , 0)
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
p( x )  15  2 x 2  x 4 is  1  0 , then p( x )    as x   .
x   :
p( x )  15  2 x 2  x 4 is 4, which is even.
p( x )    as x   , then p( x )    as x    .
The degree of
Since
The polynomial is even. Thus, the graph of p is symmetry about the y-axis.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
y

3
y
3
x

3
3
x
The Drawing of this Sketch
Since the degree of the polynomial is 4, then the graph of the polynomial can
have at most 3 local extremum points (turning points).
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
The graph of the continuous polynomial must cross the x-axis at the point
(  3 , 0 ) . Then the graph must cross the x-axis at the point ( 3 , 0 ) . In
order for this to happen, there must be a local extremum point (turning point)
3 . Since the graph of the
whose x-coordinate is between  3 and
polynomial is symmetry about the y-axis, the x-coordinate of this local
extremum point is 0. If the graph had a local extremum point (turning point)
for some x   3 , then it would have to have another one in order to pass
through the point (  3 , 0 ) . Because the graph is symmetry about the yaxis, the graph would have to have two more local extremum points (turning
points) for x  3 . Thus, the polynomial would have five local extremum
points (turning points). So, there are no local extremum points (turning
points) for x   3 nor for x  3 . The same argument shows that
there are no local extremum points (turning points) for 
for 0  x 
3  x  0 nor
3.
Thus, the graph of the polynomial has one local extremum points (turning
points).
12.
q( x )  8 x 3  14 x 4
Zeros (Roots) of q : q( x )  0  8 x 3  14 x 4  0 
2x 3 (4  7x)  0  x  0, x 
4
7
Since the factor x 3 produces the zero (root) of 0, its multiplicity is three.
Since the factor 4  7 x produces the zero (root) of
4
, its multiplicity is
7
one.
Zero (Root)
0
Multiplicity
Implication for the Graph
of the Polynomial
3
Crosses the x-axis at ( 0 , 0 )
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
4
7


Crosses the x-axis at  , 0 
4
7
1

Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
q( x )  8 x 3  14 x 4 is  14  0 , then q( x )    as x   .
x   :
q( x )  8 x 3  14 x 4
The degree of
is 4, which is even.
q( x )    as x   , then q( x )    as x    .
Since
The polynomial is neither even nor odd. Thus, the graph of the polynomial is
not symmetry with respect to the y-axis nor the origin.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
y
4
7
x
Since the degree of the polynomial is 4, then the graph of the polynomial can
have at most 3 local extremum points (turning points).
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
The graph of the continuous polynomial must cross the x-axis at the origin.


Then the graph must cross the x-axis at the point  , 0  . In order for this to
4
7

happen, there must be a local extremum point (turning point) whose xcoordinate is between 0 and
4
. If the graph had a local extremum point
7
(turning point) for some x  0 , then it would have to have another local
extremum point (turning point) in order to pass through the origin. Since
there is no symmetry in graph, this is possible and the graph would have three
local extremum points (turning points). The same argument would be true
for 0  x 
4
4
and for x  . We need calculus in order to get information
7
7
about the local extremum point(s).
Thus, the graph of the polynomial has one or three local extremum points
(turning points).
13.
f ( x )  54  27 x  2 x 3  x 4
NOTE: The expression
grouping.
54  27 x  2 x 3  x 4
can be factored by
Zeros (Roots) of f : f ( x )  0  54  27 x  2 x 3  x 4  0 
27 ( 2  x )  x 3 ( 2  x )  0  ( x  2 ) ( 27  x 3 )  0
Recalling the difference of cubes factoring formula:
a 3  b 3  (a  b)(a 2  ab  b 2 )
Thus, 27  x 3  ( 3  x ) ( 9  3 x  x 2 ) = ( 3  x ) ( x 2  3 x  9 )
Thus, ( x  2 ) ( 27  x 3 )  0  ( x  2 ) ( 3  x ) ( x 2  3 x  9 )  0 
x   2 , x  3, x 2  3x  9  0
Using the Quadratic Formula to solve x 2  3 x  9  0 , we have that
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
x 
=
b 
b 2  4a c
2a
3 
9 (1  4 )
2
=
3 
9  4 (1) 9
2
3  3 3
=
2
=
=
3 
 3  3i
2
9 [1  4 ( 1) ]
2
3
Since the factor x  2 produces the zero (root) of  2 , its multiplicity is
one. Since the factor 3  x produces the zero (root) of 3, its multiplicity is
one. Since the factor x 2  3 x  9 produces the zeros (roots) of
 3  3i 3
 3  3i 3
and
, the multiplicity of each zero (root) is
2
2
one.
Zero (Root)
Multiplicity
Implication for the Graph
of the Polynomial
2
1
Crosses the x-axis at (  2 , 0 )
3
1
Crosses the x-axis at ( 3 , 0 )
1
No implication
1
No implication
 3  3i
2
3
 3  3i
2
3
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
x  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
f ( x )    as
f ( x )  54  27 x  2 x 3  x 4 is  1  0 , then
x  .
Copyrighted by James D. Anderson, The University of Toledo
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x   :
The degree of f ( x )  54  27 x  2 x 3  x 4 is 4, which is even. Since
f ( x )    as x   , then f ( x )    as x    .
The polynomial is neither even nor odd.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as x   and as x    .
y
x
Since the degree of the polynomial is 4, then the graph of the polynomial can
have at most 3 local extremum points (turning points).
The graph of the continuous polynomial must cross the x-axis at the point
(  2 , 0 ) . Then the graph must cross the x-axis at the point ( 3 , 0 ) . In order
for this to happen, there must be a local extremum point (turning point)
whose x-coordinate is between  2 and 3. If the graph had a local
extremum point (turning point) for some x   2 , then it would have to
have another local extremum point (turning point) in order to pass through
the point (  2 , 0 ) . Since there is no symmetry in graph, this is possible and
the graph would have three local extremum points (turning points). The same
argument would be true for  2  x  3 and for x  3 . We need calculus
in order to get information about the local extremum point(s).
Copyrighted by James D. Anderson, The University of Toledo
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Thus, the graph of the polynomial has one or three local extremum points
(turning points).
14.
g ( t )  3 t 4  24 t 3  48 t 2
Zeros (Roots) of g : g ( t )  0  3 t 4  24 t 3  48 t 2  0 
3 t 2 ( t 2  8 t  16 )  0  3 t 2 ( t  4 ) 2  0  t  0 , t  4
Since the factor t 2 produces the zero (root) of 0, its multiplicity is two.
Since the factor ( t  4 ) 2 produces the zero (root) of 4, its multiplicity is
two.
Multiplicity
Implication for the Graph
of the Polynomial
0
2
Touches the t-axis at ( 0 , 0 )
4
2
Touches the t-axis at ( 4 , 0 )
Zero (Root)
Recall: If the multiplicity of the zero (root) is odd, the graph of the
polynomial crosses the x-axis, and if the multiplicity of the zero (root) is
even, the graph of the polynomial touches the x-axis.
t  :
Use the Leading Coefficient Theorem. Since the leading coefficient of
g ( t )  3 t 4  24 t 3  48 t 2 is 3 > 0, then g ( t )   as t   .
t   :
The degree of g ( t )  3 t 4  24 t 3  48 t 2 is 4, which is even. Since
g ( t )   as t   , then g ( t )   as t    .
The polynomial is neither even nor odd.
Here is the information about the zeros (roots) of the polynomial and the
graph of the polynomial as t   and as t    .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
y
y
t
t
The Drawing of this Sketch
Since the degree of the polynomial is 4, then the graph of the polynomial can
have at most 3 local extremum points (turning points).
Since the graph of the continuous polynomial touches the t-axis at the origin,
then there is a local extremum point (turning point) at the origin. Then the
graph must touch the t-axis at the point ( 4 , 0 ) . In order for this to happen,
there must be a local extremum point (turning point) whose t-coordinate is
between 0 and 4. We would need calculus in order to obtain information
about the t-coordinate of this local extremum point (turning point). Since the
graph touches at the point ( 4 , 0 ) , then there is a local extremum point
(turning point) at this point.
Thus, the graph of the polynomial has three local extremum points (turning
points).
COMMENT: It appears that the graph of the polynomial g is symmetry
about the vertical line t  2 . If we introduced a new xy coordinate system,
whose origin is at the point ( 2 , 0 ) in the ty coordinate system, we would
have that the point ( 2 , 0 )  ( t , y ) in the original ty coordinate plane is the
point ( 0 , 0 )  ( x , y ) in the new xy coordinate plane. Thus, we have that
x  t  2 , or t  x  2 .
Thus, g ( t )  3 t 4  24 t 3  48 t 2 = 3 t 2 ( t  4 ) 2 
Copyrighted by James D. Anderson, The University of Toledo
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h( x )  g ( x  2 )  3 ( x  2 ) 2 ( x  2  4 ) 2 = 3 ( x  2 ) 2 ( x  2 ) 2 =
3[ ( x  2 ) ( x  2 ) ] 2 = 3 ( x 2  4 ) 2
The polynomial h is an even function since h(  x )  3[ (  x ) 2  4 ] 2 =
3 ( x 2  4 ) 2  h( x ) Thus, the graph of the polynomial h is symmetry
about the y-axis in the xy-plane.
y
y
t,x
t  2
15.
h( x )  x 4  14 x 2  49
16.
p( x )  20 x 4  13 x 3  53 x 2  8 x  12
17.
q( x )  27 x 5  64 x 2
18.
f ( x )  150 x 3  6 x 5
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
19.
g ( x )  x 5  17 x 3  16 x
20.
h( x )  162 x  2 x 5
21.
p( t )  t 5  18 t 3  32 t
22.
f ( x )  x 5  4 x 3  8 x 2  32
23.
g( x )  9 x 4  4 x 5
24.
h( x )  8 x 5  48 x 3
Examples Find a polynomial p that has the given zeros (roots) and multiplicity.
1.
Zero (Root)
Multiplicity
4
1
2
3
1
In order for  4 to be a zero (root) of multiplicity 1, x  4 must be a
factor of p. In order for
2
3
to be a zero (root) of multiplicity 1, 3 x  2
must be a factor of p.
Thus, p ( x )  a ( x  4 ) ( 3 x  2 ) , where a is any nonzero real number.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
Since ( x  4 ) ( 3 x  2 ) = 3 x 2  2 x  12 x  8 = 3 x 2  10 x  8 ,
then p( x )  a ( 3 x 2  10 x  8 )
Answer: p( x )  a ( 3 x 2  10 x  8 ) , where a is any nonzero real
number
2.
Zero (Root)
Multiplicity
3
1
5
1
5
1
In order for  3 to be a zero (root) of multiplicity 1, x  3 must be a
factor of p. In order for  5 to be a zero (root) of multiplicity 1, x  5
must be a factor of p. In order for 5 to be a zero (root) of multiplicity 1,
x  5 must be a factor of p.
Thus, p ( x )  a ( x  3 ) ( x  5 ) ( x  5 ) , where a is any nonzero real
number.
Using the special product formula ( a  b ) ( a  b )  a 2  b 2 , we have
that ( x  5 ) ( x  5 ) = x 2  25 .
Thus, ( x  3 ) ( x  5 ) ( x  5 ) = ( x  3 ) ( x 2  25 ) =
x 3  25 x  3 x 2  75 = x 3  3 x 2  25 x  75
Thus, p( x )  a ( x 3  3 x 2  25 x  75 )
Answer: p( x )  a ( x 3  3 x 2  25 x  75 ) , where a is any nonzero real
number
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
3.
Zero (Root)
Multiplicity
2
1
4
2
In order for 2 to be a zero (root) of multiplicity 1, x  2 must be a factor
of p. In order for 4 to be a zero (root) of multiplicity 2, ( x  4 ) 2 must be
a factor of p.
Thus, p ( x )  a ( x  2 ) ( x  4 ) 2 , where a is any nonzero real number.
Using the special product formula ( a  b ) 2  a 2  2 a b  b 2 , we have
that ( x  4 ) 2  x 2  8 x  16 .
Thus, ( x  2 ) ( x  4 ) 2 = ( x  2 ) ( x 2  8 x  16 ) =
x 3  8 x 2  16 x  2 x 2  16 x  32 = x 3  10 x 2  32 x  32
Thus, p( x )  a ( x 3  10 x 2  32 x  32 )
Answer: p( x )  a ( x 3  10 x 2  32 x  32 ) , where a is any nonzero
real number
4.
Zero (Root)
Multiplicity
1
1
6
1
 7i
1
7i
1
In order for  1 to be a zero (root) of multiplicity 1, x  1 must be a factor
of p. In order for 6 to be a zero (root) of multiplicity 1, x  6 must be a
Copyrighted by James D. Anderson, The University of Toledo
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factor of p. In order for  7 i to be a zero (root) of multiplicity 1, x  7 i
must be a factor of p. In order for 7 i to be a zero (root) of multiplicity 1,
x  7 i must be a factor of p.
Thus, p ( x )  a ( x  1 ) ( x  6 ) ( x  7 i ) ( x  7 i ) , where
nonzero real number.
a
is any
We will use the special product formula ( a  b ) ( a  b )  a 2  b 2 for
( x  7 i ) ( x  7 i ) . Thus, ( x  7 i ) ( x  7 i )  x 2  49 i 2 . Since i   1 ,
then i 2   1 . Thus, ( x  7 i ) ( x  7 i )  x 2  49 i 2  x 2  49 .
Since ( x  1) ( x  6 )  x 2  6 x  x  6  x 2  5 x  6 , then
( x  1 ) ( x  6 ) ( x  7 i ) ( x  7 i ) = ( x 2  5 x  6 ) ( x 2  49 ) =
( x 2  49 ) ( x 2  5 x  6 ) = x 4  5 x 3  6 x 2  49 x 2  245 x  294 =
x 4  5 x 3  43 x 2  245 x  294
Thus, p( x )  a ( x 4  5 x 3  43 x 2  245 x  294 )
Answer: p( x )  a ( x 4  5 x 3  43 x 2  245 x  294 ) , where a is any
nonzero real number
5.
Zero (Root)
Multiplicity
5
4
2
2  3i
1
2  3i
1

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
In order for

5
4
to be a zero (root) of multiplicity 2, ( 4 x  5 ) 2 must be a
factor of p. In order for 2  3 i to be a zero (root) of multiplicity 1,
x  ( 2  3 i ) must be a factor of p. In order for 2  3 i to be a zero
(root) of multiplicity 1, x  ( 2  3 i ) must be a factor of p.
Thus, p ( x )  a ( 4 x  5 ) 2 [ x  ( 2  3i ) ][ x  ( 2  3i ) ] , where a is
any nonzero real number.
Since [ x  ( 2  3 i ) ][ x  ( 2  3 i ) ] = [ ( x  2 )  3 i ][ ( x  2 )  3 i ] ,
then using the special product formula ( a  b ) ( a  b )  a 2  b 2 , we
2
2
have that [ ( x  2 )  3 i ][ ( x  2 )  3 i ] = ( x  2 )  9 i . Using the
( a  b ) 2  a 2  2 a b  b 2 , we have that
special product formula
( x  2 ) 2  x 2  4 x  4 . Thus, ( x  2 ) 2  9 i 2 = x 2  4 x  4  9
= x 2  4 x  13 .
Using the special product formula ( a  b ) 2  a 2  2 a b  b 2 , we have
that ( 4 x  5 ) 2  16 x 2  40 x  25 .
Thus, ( 4 x  5 ) 2 [ x  ( 2  3i ) ][ x  ( 2  3i ) ] =
( 16 x 2  40 x  25 ) ( x 2  4 x  13 ) = 16 x 4  64 x 3  208 x 2  40 x 3
 160 x 2  520 x  25 x 2  100 x  325 =
16 x 4  24 x 3  73 x 2  420 x  325
Thus, p( x )  a ( 16 x 4  24 x 3  73 x 2  420 x  325 )
Answer: p( x )  a ( 16 x 4  24 x 3  73 x 2  420 x  325 ) , where a is
any nonzero real number
6.
Zero (Root)
8
Multiplicity
5
Copyrighted by James D. Anderson, The University of Toledo
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1
2
3
11
7
4
6
2

In order for  8 to be a zero (root) of multiplicity 5, ( x  8 ) 5 must be a
factor of
p.
In order for

1
2
to be a zero (root) of multiplicity 3,
( 2 x  1 ) 3 must be a factor of p. In order for 11 to be a zero (root) of
7
multiplicity 4, ( 7 x  11) must be a factor of p. In order for 6 to be a
zero (root) of multiplicity 2, ( x  6 ) 2 must be a factor of p.
4
Thus, p ( x )  a ( x  8 ) 5 ( 2 x  1) 3 ( 7 x  11) 4 ( x  6 ) 2 , where a is
any nonzero real number.
Answer: p ( x )  a ( x  8 ) 5 ( 2 x  1) 3 ( 7 x  11) 4 ( x  6 ) 2 , where a is
any nonzero real number
7.
Zero (Root)
Multiplicity
7
3
1
7
3
1

i
5
1
i
5
1
In order for

7
3
to be a zero (root) of multiplicity 1, 3 x  7 must be a
factor of p. In order for
7
3
to be a zero (root) of multiplicity 1, 3 x  7
Copyrighted by James D. Anderson, The University of Toledo
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In order for
i
5
must be a factor of
p.
multiplicity 1, x  i
5 must be a factor of p. In order for i
zero (root) of multiplicity 1, x  i
to be a zero (root) of
5 to be a
5 must be a factor of p.
Thus, p ( x )  a ( 3 x  7 ) ( 3 x  7 ) ( x  i
any nonzero real number.
5 )( x  i
5 ) , where a is
We will use the special product formula ( a  b ) ( a  b )  a 2  b 2 for
( 3 x  7 ) ( 3 x  7 ) . Thus, ( 3 x  7 ) ( 3 x  7 )  9 x 2  49 .
We will use the special product formula ( a  b ) ( a  b )  a 2  b 2 for
( x  i 5 ) ( x  i 5 ) . Thus, ( x  i 5 ) ( x  i 5 )  x 2  5i 2 . Thus,
( x  i 5 )( x  i
5 )  x 2  5i 2  x 2  5 .
Thus, ( 3 x  7 ) ( 3 x  7 ) ( x  i
5 )( x  i
5 ) = ( 9 x 2  49 ) ( x 2  5 ) =
9 x 4  45 x 2  49 x 2  245 = 9 x 4  4 x 2  245
Thus, p( x )  a ( 9 x 4  4 x 2  245 )
Answer:
number
8.
p( x )  a ( 9 x 4  4 x 2  2 4 5) , where a is any nonzero real
Zero (Root)
Multiplicity
3
2
3
2
i
2
i
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320
In order for  3 to be a zero (root) of multiplicity 2, ( x  3 ) 2 must be a
factor of p. In order for 3 to be a zero (root) of multiplicity 2, ( x  3 ) 2
must be a factor of p. In order for  i to be a zero (root) of multiplicity 2,
( x  i ) 2 must be a factor of p. In order for i to be a zero (root) of
multiplicity 2, ( x  i ) 2 must be a factor of p.
Thus, p ( x )  a ( x  3 ) 2 ( x  3 ) 2 ( x  i ) 2 ( x  i ) 2 , where a is any
nonzero real number.
Since ( a b ) n  a n b n , then ( x  3 ) 2 ( x  3 ) 2 = [ ( x  3 ) ( x  3 ) ] 2 =
( x 2  9 ) 2 = x 4  18 x 2  81 and ( x  i ) 2 ( x  i ) 2 =
[ ( x  i ) ( x  i ) ] 2 = ( x 2  1) 2 = x 4  2 x 2  1
Thus, ( x  3 ) 2 ( x  3 ) 2 ( x  i ) 2 ( x  i ) 2 =
( x 4  18 x 2  81) ( x 4  2 x 2  1) = x 8  2 x 6  x 4  18 x 6  36 x 4
 18 x 2  81 x 4  162 x 2  81 = x 8  16 x 6  46 x 4  144 x 2  81
Thus, p( x )  a ( x 8  16 x 6  46 x 4  144 x 2  81)
Answer: p( x )  a ( x 8  16 x 6  46 x 4  144 x 2  81) , where a is any
nonzero real number
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1320